Q. 9.56 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the final concentration of each of the following:

a. $1.0 \mathrm{L} \mathrm{of} \mathrm{a} 4.0 \mathrm{M} {\mathrm{HNO}}_3$ solutions is added to the water so that the final volume is $8.0 \mathrm{L}$.

b. Water is added to $0.25 \mathrm{L} \mathrm{of} \mathrm{a} 6.0 \mathrm{M} \mathrm{NaF}$ the solution to make $2.0 \mathrm{L}$ of a diluted $\mathrm{NaF}$ solution.

c. A $50.0‐\mathrm{mL}$ sample of a $8.0 \% (\mathrm{m}/\mathrm{v}) \mathrm{KBr}$ solution is diluted with water so that the final volume is $200.0 \mathrm{mL}$.

d. A $5.0‐\mathrm{mL}$ sample of an $50.0 \% (\mathrm{m}/\mathrm{v})$ acetic acid $\left({\mathrm{HC}}_2{\mathrm{H}}_3{\mathrm{O}}_2\right)$ solution is added to water to give a final volume of $25 \mathrm{mL}$.

Step-by-Step Solution

Answer

a. The final concentration of $1.0 \mathrm{L} \mathrm{of} \mathrm{a} 4.0 \mathrm{M} {\mathrm{HNO}}_3$solutions is added to the water so that the final volume is $8 \mathrm{L}$ is $0.5 \mathrm{M}$.

b. The final concentration when water is added to $0.25 \mathrm{L} \mathrm{of} \mathrm{a} 6.0 \mathrm{M} \mathrm{NaF}$  the solution to make  $2.0 \mathrm{L}$ of a diluted $\mathrm{NaF}$ solution is $0.75 \mathrm{M}$.

c. The final concentration of $50.0‐\mathrm{mL}$  the sample of a $8.0 \% (\mathrm{m}/\mathrm{v}) \mathrm{KBr}$ solution is diluted with water so that the final volume is $200.0 \mathrm{mL}$ is $2 \%$.

d. The final concentration of $5.0‐\mathrm{mL}$  sample of an $50.0 \% (\mathrm{m}/\mathrm{v})$ acetic acid $\left({\mathrm{HC}}_2{\mathrm{H}}_3{\mathrm{O}}_2\right)$ solution is added to water to give a final volume of $25 \mathrm{mL}$ is $10\%$.

Part (a) step 1: Given Information

We need to calculate the final concentration of the $1.0 \mathrm{L} \mathrm{of} \mathrm{a} 4.0 \mathrm{M} {\mathrm{HNO}}_3$ solutions is added to the water so that the final volume is$8.0 \mathrm{L}$.  

Part (a) step 2: Explanation

We know

molarity $\left(\mathrm{M}\right)$ $= 4\mathrm{M}$

Litre is solution $= 1 \mathrm{L}$

The moles of solute = $1\times 4=4\text{moles}$

The final volume of solution $= 8 \mathrm{L}$

Molarity $\left(\mathrm{M}\right)=\frac{4 \mathrm{mole} }{8 \mathrm{L}}=0.5 \mathrm{M}$

Part (b) step 1: Given Information

We need to calculate the final concentration when water is added to $0.25 \mathrm{L} \mathrm{of} \mathrm{a} 6.0 \mathrm{M} \mathrm{NaF}$the solution to make $2.0 \mathrm{L}$ of a diluted $\mathrm{NaF}$ solution.

Part (b) step 2: Explanation

We know,

molarity $\left(\mathrm{M}\right)=6 \mathrm{M}$

Litre is solution $=0.25 \mathrm{L}$

The moles of solute = $6\times 0.25=1.5\text{moles}$

The final volume of solution $=2 \mathrm{L}$

Molarity $\left(\mathrm{M}\right)=\frac{1.5 \mathrm{mole} }{6 \mathrm{L}}=0.75 \mathrm{M}$

Part (c) step 1: Given Information

We need to calculate the final concentration of $50.0‐\mathrm{mL}$  sample of a $8.0 \% (\mathrm{m}/\mathrm{v}) \mathrm{KBr}$ solution is diluted with water so that the final volume is $200.0 \mathrm{mL}$. 

Part (c) step 2: Explanation

We know,

initial volume $=50 \mathrm{mL} = 0.05\mathrm{L}$

$$\begin{gathered} \% \mathrm{mass} (\mathrm{m}/\mathrm{v}) = 8\% \\ \mathrm{mass} \mathrm{of} \mathrm{solute}=\frac{\% \mathrm{mass}}{100}\times \left(\mathrm{solute} \mathrm{volume}\right) \\ =\frac{8}{100}\times 0.05=00.4 \mathrm{g} \end{gathered}$$

Final volume of solution $=200 mL = 0.2 L$

New $\% \mathrm{mass}(\mathrm{m}/\mathrm{v}) =\frac{0.004 \mathrm{g}}{0.2}\times 100=2\%$

Part (d) step 1: Given Information

We need to calculate the final concentration of  $5.0‐\mathrm{mL}$ the sample of an $50.0 \% (\mathrm{m}/\mathrm{v})$
 an acetic acid $\left({\mathrm{HC}}_2{\mathrm{H}}_3{\mathrm{O}}_2\right)$ solution is added to water to give a final volume of $25 \mathrm{mL}$ .

Part (d) step 2: Explanation

We know,

initial volume $=5 \mathrm{mL} = 0.005 \mathrm{L}$

$$\begin{gathered} \% \mathrm{mass} (\mathrm{m}/\mathrm{v}) = 50\% \\ \mathrm{mass} \mathrm{of} \mathrm{solute} =\frac{\% \mathrm{mass}}{100}\times \left(\mathrm{solute} \mathrm{volume}\right) \\ =\frac{50}{100}\times 0.005=0.0025 \mathrm{g} \end{gathered}$$

Final volume of solution is $=25 \mathrm{mL} = 0.025 \mathrm{L}$

New, $\% \mathrm{mass} (\mathrm{m}/\mathrm{v}) =\frac{0.0025 \mathrm{g}}{0.025}\times 100 = 10\%$